Question

A spherically-symmetric charge distribution takes the form... Read on the attachment below

Answer 1

use spherical coordinate to integrate.

Answer 2

if you can do the a for me? plz

Answer 3

Using Gauss' Law,
$$
\Large
\oint\mathbf{E}\cdot\mathrm{d}\mathbf{S} = \frac{1}{\varepsilon_0} \iiint_\Omega \rho (r)\,\mathrm{d}V
$$
Where the 1st integral is taken over a closed surface. E is the electric field and S is the surface.

Now, the second integral is the sum of all charges within this closed surface divided by the permittivity constant, \(\varepsilon_0\)
For \(r\le R\),
$$
\large{
\rho= A\left(1-\cfrac{r}{R}\right )\\
\frac{1}{\varepsilon_0} \iiint_\Omega A\left(1-\cfrac{r}{R}\right )\,\mathrm{d}V
}
$$
Using spherical coordinates, this becomes,
$$\large{
\frac{1}{\varepsilon_0} \int_{\varphi=0}^{2 \pi} \int_{\theta=0}^{\pi} \int_{r=0}^{R} A\left(1-\cfrac{r}{R}\right ) r^2 \sin \theta \,\mathrm{d}r\ \mathrm{d}\theta\ \mathrm{d}\varphi\\
= \frac{2\pi}{\varepsilon_0} \int_{\theta=0}^{\pi} \int_{r=0}^{R} A\left(1-\cfrac{r}{R}\right ) r^2 \sin \theta \,\mathrm{d}r\ \mathrm{d}\theta\ \\
=\frac{4\pi}{\varepsilon_0}\int_{r=0}^{R} A\left(1-\cfrac{r}{R}\right ) r^2 \,\mathrm{d}r\\
=\frac{4\pi}{\varepsilon_0}\int_{r=0}^{R} A\left(r^2-\cfrac{r^3}{R}\right ) \,\mathrm{d}r\\
}
$$
Which can easily be evaluated. This gives the answer for a) the total charge Q when multiplied by \(\varepsilon_0 \).
The case for \(r\ge R\) is computed similarly.